# Exponentiation of Center of Lie Algebra

Let $G$ be a Lie Group, and $g$ its Lie Algebra. Show that the subgroup generated by exponentiating the center of $g$ generates the connected component of $Z(G)$, the center of $G$.

Source: Fulton-Harris, Exercise 9.1

The difficulty lies in showing that exponentiating the center of $g$ lands in the center of $G$. Since the image of $exp(Z(g))$ is the union of one parameter subgroups that are disjoint, we know it connected. Also I can show this for the case when $G = Aut(V)$ and $g = End(V)$ since we have $G \subset g$.

EDIT: G not connected

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$G$ is connected and so is generated by elements of the form $\exp Y$ for $Y \in g$. Therefore it is sufficient to show that for $X \in Z(g)$ $\exp X$ and $\exp Y$ commute. Now define $\gamma: \mathbb R \to G$ by $\gamma(t) = \exp(X)\exp(tY)\exp(-X)$. Then $$\gamma'(0) = Ad_{\exp(X)} Y = e^{ad_X} Y = (1 + ad_X + \frac{1}{2}ad_X \circ ad_X + \cdots)(Y) = Y.$$ But it is also easily seen that $\gamma$ is a homomorphism, i.e. $\gamma(t+s) = \gamma(t)\gamma(s)$. This characterizes the exponential map so that $\gamma(t) = \exp(tY)$. Taking $t=1$ shows that $\exp X$ and $\exp Y$ commute.
I'm confused. Take $O(2)$ for example. The lie algebra is the Lie algebra of $SO(2)$ which is abelian. Thus exponentiating the center gives all of $SO(2)$, but this is not the identity component of the center of $O(2)$ (which has size 2). Also, what last statement is bothering you? $F: t\mapsto \exp(tY)$ is characterized by $F(t+s) = F(t)F(s)$ and $F'(0) = Y$. Since $\gamma$ also satisfies these we have $\exp(tY) = \gamma(t) = \exp(X) \exp(tY) \exp(-X)$. This says $\exp(tY)$ and $\exp(X)$ commute for all $t$. – Eric O. Korman Jul 13 '12 at 2:55
I don't think your answer is correct after speaking to Prof. Harris. For example, the dimension of the lie algebra of $SO(2)$ is $1$, so its exponentiation should have dimension $1$, and isn't all of $SO(2)$, it is the center of $SO(2)$ (since $SO(2)$ is connected). I think the original problem stands, $G$ need not be connected. – bijection Jul 17 '12 at 16:00